Let us consider the representation theory of $SU(2)$. There is a unique irreducible representation of dimension $n$ for each $n \ge 1$, which we will denote $\mathbf{n}$, with the defining $2$-dimensional representation $\mathbf{2}$ being called the fundamental representation. It is well known that all the irreducible representations can be built up from tensor powers of the fundamental representation. Moreover, from the Schur-Weyl duality, we have ways of classifying the symmetries of the tensor representations.

For example, the tensor power $\mathbf{2}\otimes \mathbf{2}$ decomposes as $$\mathbf{2}\otimes \mathbf{2} = \mathbf{1} \oplus \mathbf{3},$$ where $\mathbf{1} \simeq\Lambda^2(\mathbf{2})$ is the space of alternating tensors over $\mathbf{2}$ and where $\mathbf{3} \simeq S^2(\mathbf{2})$ is the space of symmetric tensors over $\mathbf{2}$.

Likewise, for higher tensor powers of the fundamental representation, we can decompose the space into irreducible representations composed of tensors of mixed symmetry, corresponding to different Young tableaux.

As far as I'm aware, the techniques that I know of only allow us to do this sort of decomposition for tensor powers of the fundamental rep. Is there anyway to generalize this to other irreducible reps? For example, we know that $$\mathbf{3} \otimes \mathbf{3} = \mathbf{1} \oplus \mathbf{3} \oplus \mathbf{5}.$$ Do these irreducible subspaces correspond to symmetric/antisymmetric tensors over $\mathbf{3}$? If so, how can we find the relevant decomposition? Are there ways of classifying the symmetric/anti-symmetric/mixed-symmetric subspaces of $\mathbf{n}^{\otimes k}$ in general?

I am looking for references relating to the above questions (although if the answer is short and simple, feel free to just answer the question). I don't know very much representation theory, so I don't know if this is a well known problem with a solution or if it's intractable. If there are solutions for $SU(N)$ in general, I would be interested to know as well.

  • $\begingroup$ My Lie group knowledge is somewhat lacking here, but: How closely related arr these reps to the reps of the complexification of the Lie algebra? Because that is $\mathfrak{sl}_2$ which has precisely the same property of one irrep for each dimension, and here the tensor products are well-known (essentially given by the Littlewood-Richardson rule). $\endgroup$ Commented Jun 18, 2016 at 19:08
  • $\begingroup$ @TobiasKildetoft We can consider the complexification (and perhaps that's more natural), nothing should change. The tensor product decomposition is indeed well known, and given in general by the Littlewood-Richardson rule. But I'm not after just the tensor product decomposition itself, but a decomposition in terms of symmetry. To give another example of the kind of thing I'm after, consider $\mathbf{2}\otimes\mathbf{2}\otimes\mathbf{2}$. The Littlewood-Richardson rule tells us that $\mathbf{2}\otimes \mathbf{2} \otimes \mathbf{2} = \mathbf{4} \oplus \mathbf{2} \oplus \mathbf{2}$. $\endgroup$
    – EuYu
    Commented Jun 18, 2016 at 19:43
  • $\begingroup$ I am not sure what you mean by a decomposition in terms of symmetry. What symmetry? $\endgroup$ Commented Jun 18, 2016 at 19:44
  • $\begingroup$ But a more detailed analysis would show that $\mathbf{4}$ is the subspace of tensors fully symmetric in the three copies of $\mathbf{2}$, and the remaining copies of $\mathbf{2}$ are symmetric/antisymmetric respectively in the first two copies. A decomposition like that for, say $\mathbf{3}\otimes\mathbf{3}\otimes\mathbf{3}$ would be something I'm looking for. $\endgroup$
    – EuYu
    Commented Jun 18, 2016 at 19:46
  • 1
    $\begingroup$ The symmetries of all Kronecker compositions are utilized in here. $\endgroup$ Commented Jan 17, 2018 at 16:07

2 Answers 2


This is a question that touches on many issues. On the one hand, things are indeed easier to deal with in the language of the complexification. For $SL(2,\mathbb{C})$, there is a specific element $H$ in the Lie algebra $\mathfrak{sl}(2,\mathbb{C})$, which acts diagonalizably on any finite dimensional representation. The eigenvalues are called weights, and in the case of an irreducible representation, they are all different. Specifically, the $n$-dimensional irreducible representation $\mathbf{n}$ admits a basis of weight vectors of weight $-n+1,-n+3,\dots,n-3,n-1$. Now in a tensor product, the tensor product of two eigenvectors for $H$ is an eigenvector with eigenvalue the sum of the eigenvalues of the two factors. There is a general result saying that the weight vector with highest possible weight will always generate an irreducible subrepresentation, and there there is an invariant complement to this subrepresentation. This allows you to compute decompositions in an elementary way.

Take the example of $\mathbf{3} \otimes \mathbf{3}$. In $\mathbf{3}$ you have weights $-2$, $0$, and $2$, so in the tensor product, the possible weights are $-4$, $-2$, $0$, $2$, and $4$ and the dimensions of the eigenspaces are $1$, $2$, $3$, $2$, and $1$. The weight vector of weight $4$ generates a subrepresentation $\mathbf{5}$, which has one weight vector for each of the listed weights. So in the complement, you get weights $-2$, $0$, and $2$ with dimensions $1$, $2$, and $1$. The highest of these gives you a subrepresentation $\mathbf{1}$ again including one of each of the weights, so there is just one copy of $\mathbf{0}$ left. To describe the result in terms of symmetry, you observe that the weight vector of the maximal weight $4$ is the tensor product of a highest weight vector with itself, so this sits in $S^2\mathbf{3}$. Either counting dimensions or by direct analysis of the weights you can see that $S^2\mathbf{3} \cong \mathbf{5} \oplus \mathbf{1}$ and $\Lambda^2\mathbf{3} \cong \mathbf{3}$.

This works similarly for general tensor products $\mathbf{n} \otimes \mathbf{m}$ for $n \geq m$. This is isomorphic to $\mathbf{n+m-1} \oplus \dots \oplus \mathbf{n-m+1}$ with dimension decreasing by two in each step (so there are always $m$ summands). For $n=m$, things look similar as in the special case, $S^2\mathbf{n} = \mathbf{2n-1} \oplus \mathbf{2n-5} \oplus \dots$ and $\Lambda^2\mathbf{n} = \mathbf{2n-3} \oplus \mathbf{2n-7} \oplus \dots$ with dimensions going down in steps of $4$.

You can always construct higher tensor powers step by step, say $\mathbf{2}\otimes \mathbf{2} \otimes \mathbf{2} \cong (\mathbf{3} \oplus \mathbf{1})\otimes \mathbf{3} \cong (\mathbf{3} \otimes \mathbf{2}) \oplus \mathbf{2}$ and then the first summand splits as $\mathbf{4} \oplus \mathbf{2}$. However, this is just one possible way to “decompose according to symmetry”, since one has distinguished the first two factors. In fact, the canonical way to decompose higher tensor products is just into so-called isotypical components. Here this would be $\mathbf{2} \otimes \mathbf{2} \otimes \mathbf{2} \cong \mathbf{4} \oplus W$, where $W$ is an invariant subspace isomorphic to a direct sum of two copies of $\mathbf{2}$. However, there are various possible realizations for this isomorphism, none of which is canonical. the systematic way to deal with this is simultaneously decomposing as a representation of $\mathfrak{sl}(2,\mathbb{C})$ and the permutation group $\mathfrak{S}_3$. This indeed leads towards Young diagrams, which are also needed to deal with $SL(n,\mathbb{C})$ and hence with $SU(n)$. If you are looking for literature in that direction, I would recommend the book of Fulton and Harris on representation theory.

  • $\begingroup$ Thank you for your detailed answer Andreas. The middle example of yours is exactly the type of thing I am looking for. Suppose we consider $\mathbf{3}\otimes\mathbf{3}$. On the one hand, from the weight decomposition, we know that this is $\mathbf{1}\oplus\mathbf{3}\oplus\mathbf{5}$. On the other hand, we also know that $\mathbf{3}\otimes\mathbf{3} \simeq S^2(\mathbf{3})\oplus \Lambda(\mathbf{3})$. As you've said, from dimensionality arguments, we can see that for this simple case, we have $S^2(\mathbf{3}) \simeq \mathbf{1} \oplus \mathbf{5}$ and $\mathbf{3} \simeq \Lambda(\mathbf{3})$. $\endgroup$
    – EuYu
    Commented Jun 20, 2016 at 2:14
  • $\begingroup$ But is there a way to do this in full generality? For example, say we consider $\mathbf{3}\otimes\mathbf{3}\otimes \mathbf{3}$. Then we have a decomposition into a rather large number of irreducible subreps, which we should be able to match onto the various symmetric subspaces corresponding to the different Young symmetrizers. In the case of tensor powers of the fundamental rep, we are guaranteed that each of these symmetric subspaces are irreducible themselves, but this no longer holds for $n>2$. How do we know which irreps belong to which symmetric subspaces in this case? $\endgroup$
    – EuYu
    Commented Jun 20, 2016 at 2:19
  • $\begingroup$ I don't think that there is a general answer for this. Of course, $\Lambda^3\textbf{3}=\textbf{1}$ and it turns out that $S^3\textbf{3}=\textbf{7}\oplus \textbf{3}$. But there are two more copies of $\textbf{3}$ in $\textbf{3}\otimes\textbf{3}\otimes\textbf{3}$, which have to be contained in the two remaining (isomorphic) symmetry types, but similarly as indicated in my answer, only the sum of the two symmetry types is canonical, a decomposition into two summands involves a choice. $\endgroup$ Commented Jun 20, 2016 at 14:07
  • $\begingroup$ @AndreasCap: It's not well known but there is in fact a general answer, at least for $SL_2$. See my answer to this post. $\endgroup$ Commented Jun 2, 2020 at 18:16

Good question! The answer, known as the Cayley-Sylvester formula, has been figured out in the 1850's. First let me switch notations from $\mathbf{n}$ to $S^{n-1}\mathbf{2}$, for the $n$-dimensional irreducible representation of $SU(2)$ realized as the $(n-1)$-th symmetric power of the fundamental two-dimensional one. In general, one has $$ S^m(S^n\mathbf{2})=\oplus_{w=0}^{\lfloor\frac{mn}{2}\rfloor} M_{m,n,w}\ S^{mn-2w}\mathbf{2} $$ where the multiplicities $M_{m,n,w}$ are given by $$ M_{m,n,w}=[m,n,w]-[m,n,w-1]\ . $$ Here, $[m,n,w]$ denotes the number of integer partitions of weight $w$ with at most $m$ parts of length at most $n$. By convention $[m,n,w]=0$ if $w<0$.

For the totally antisymmetric part, one can reduce to the previous situation via the isomorphism $$ \wedge^m S^n\mathbf{2}=S^m(S^{n-m+1}\mathbf{2})\ . $$ For more on this Wronskian isomorphism, see my article with Chipalkatti page 46. For example, the symmetric part of $\mathbf{3}\otimes\mathbf{3}\otimes\mathbf{3}$, i.e., $S^{3}S^2\mathbf{2}$ can be computed as follows. One has $[3,2,0]=1$ (the empty partition), $[3,2,1]=1$ (the partition (1)), $[3,2,1]=2$ (the partitions (2) and (1,1)), and $[3,2,3]=2$ (the partitions (2,1) and (1,1,1)). Therefore $M_{3,2,0}=1$, $M_{3,2,1}=0$, $M_{3,2,2}=1$ and $M_{3,2,0}=0$. So $$ S^{3}\mathbf{3}=\mathbf{7}\oplus\mathbf{3}\ . $$

The formula for the decomposition of $S^m(S^n\mathbf{2})$ was discovered by Cayley in a letter from around 1854 starting by "Dear Sylvester, Eureka." However he left a substantial gap in the derivation and the first complete proof is due to Sylvester in an 1878 article. This is also the first proof of unimodality of $q$-binomial coefficients. A nice account of the history around the Cayley-Sylvester formula is in the article "The rise of Cayley's invariant theory (1841–1862)" by T. Crilly.

I don't know if one has similar nice formulas for mixed symmetry. There is a formula for $s_{\lambda}(1,q,q^2,\ldots)$ due to Stanley, but I am not sure it is easy to extract the multiplicities from it. The formula (Stanley's hook-content formula) is a $q$-analogue of the hook length formula and says $$ s_{\lambda}(1,q,q^2,\ldots,q^{m-1})=q^{\sum_i (i-1)\lambda_i} \prod_{\square\in\lambda}\frac{1-q^{m+c(\square)}}{1-q^{h(\square)}} $$ where $h(\square)$ is the hook and $c(\square)$ is the content. In the $SU(N)$ case, some examples are known but in the general this a major open problem in representation theory called the plethysm problem.


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